Brief answers to the homework problems in MAT 410: Introduction to Topology =========================================================================== Stefan Kohl 1. The supremum is 1 (calculus). 2. 1. Geometric series, limit 1. 2. The polynomials in numerator and denominator have the same degree and the same leading coefficient, so the limit is 1. 3. Any enumeration of the rationals in [0,1] will do -- e.g. 0, 1, 0, 1/2, 1, 0, 1/3, 2/3, 1, 0, 1/4, 2/4, 3/4, 1, 0, 1/5, 2/5, 3/5, 4/5, 1, ... . Repetitions don't matter. 4. 1. Q is dense, so the set of limit points is R. 2. The set is dense in R_0^+ (nonnegative real numbers), so the set of limit points is R_0^+. 3. The zeros of sin(1/x) are the numbers 1/(k*pi) for k in Z, so the set of limit points is {0}. 5. The characteristic function of any dense set whose complement is also dense, hence e.g. of Q, will do. 6. E.g. f: [0,1[ -> R, x |-> 1/(1-x). 7. Let U be the area of the union and I be the area of the intersection. Put A_1 := pi*r_1^2 and A_2 := pi*r_2^2. W.o.l.g. assume r_1 >= r_2. - If the balls are disjoint, U = A_1 + A_2 and I = 0. - If the second ball is contained in the first, U = A_1 and I = A_2. - If the balls intersect nontrivially: - Put d := sqrt((x_1 - y_1)^2 + (x_2 - y_2)^2). - Put a := arccos((d^2 + r_1^2 - r_2^2)/(2*d*r_1)). - Put h := r_1 sin(a) and b := arcsin(h/r_2). - Put h' := sqrt(r_1^2 - h^2) and h" := sqrt(r_2^2 - h^2). - Put S_1 := a * r_1^2 and S_2 := b * r_2^2. - Put T_1 := h*h' and T_2 := h*h". - Put D := S_2 + T_1 - S_1 - T_2. Now, U = A_1 + D and I = A_2 - D. 8. Let d(x,y) = 0 if x=y and d(x,y) = 1 if x<>y (discrete metric). 9. 1. See answer to Exercise 8, 2. Euclidean distance function, 3. Manhattan distance function (sum of coordinate differences). 10. 1. All sets are both open and closed. 2. A sequence in a discrete metric space converges iff it eventually gets constant. 11. Due to the Axiom of Symmetry, a distance function on X is defined by d(x,y), d(x,z) and d(y,z). Each of these distances can take value 1,2 or 3. There are 3^3=27 possibilities, of which only (1,1,3), (1,3,1) and (3,1,1) are ruled out by the Triangle Inequality. So there are 27-3=24 possible distance functions, which are parametrized by the remaining triples. 12. 1. Always. -- Since all sets are open, preimages of open sets are always open. 2. If and only if it is bijective. 13. It suffices to show that the metrics are equivalent, i.e. that the values of the distance functions differ by at most a constant factor. This is handled as case L=2 in the answer to Exercise 14. 14. It suffices to show that the metrics are equivalent, i.e. that d_1(x,y) and d_L(x,y) differ by at most a constant factor. We write "sum" for the summation over i from 1 to n. (1) d_1(x,y) = sum |x_i-y_i| >= max|x_i-y_i| >= 1/n * (sum |x_i-y_i|^L)^(1/L) = 1/n * d_L(x,y). (2) d_L(x,y) = (sum |x_i-y_i|^L)^(1/L) >= max|x_i-y_i| >= 1/n * sum |x_i-y_i| = 1/n * d_1(x,y). So d_1(x,y) and d_L(x,y) differ by not more than a factor of n, hence are equivalent as required. 15. The discrete metric works: d(x,y) = 0 if x=y, and d(x,y) = 1 if x<>y. 16: A nicely worked out solution can be found at http://www.gap-system.org/~john/MT4522/Solutions/S5.7.html. 17. The answer is "no" -- the sequence (a_n) of 2x2 matrices with 1's on the main diagonal, 0 below and n above is an example of a sequence of elements of SL(2,R) which has no convergent subsequence. 18. Take the trivial topology (i.e. the only open sets are the empty set and X itself). Then every sequence of points of X converges to every point in X. 19. The axioms for a topological space are satisfied: - The empty set and Z are open by definition. - A union of unions of residue classes is again a union of residue classes, so unions of open sets are open. - The intersection of two residue classes is a residue class or empty, hence finite intersections of unions of residue classes are also unions of residue classes. So, finite intersections of open sets are open. We have a Hausdorff space, since given two points a and b, for any positive integer m which does not divide a-b, a+mZ and b+mZ are disjoint open sets containing a and b, respectively. 20. Residue classes are both open and closed, so if there would be only finitely many primes, the union of the residue classes pZ taken over all primes would be closed. Then the complement of this union would be open. However, since the only integers which are not integer multiples of primes are -1 and 1, this complement is {-1,1}, which is finite and hence certainly not a union of residue classes. This contradiction shows that there are infinitely many primes. [This proof is known as "Fuerstenberg's proof", named after Harry (Hillel) Fuerstenberg.] 21. For example, n -> n+1 and n -> -n are homeomorphisms of the topology in question, since they map residue classes to residue classes. 22. The closure is R itself, since it is a closed set. The interior is empty, since R does not contain a ball of radius > 0 in C. 23. 1. The closure is R, since Q is dense in R. The interior is S itself, since S is a union of open sets, hence open. 2. The closure is R, since S is infinite and R is the only infinite closed set in the Zariski topology. Now, S is a union of intervals of length 1/2^i for i from 0 to infinity. The sum of the lengths of these intervals is 2, so the complement of S in R is infinite. Hence the interior is empty, since the only open set with infinite complement in the Zariski topology is the empty set. 24. The open sets in the product topology on X_1 x X_2 are the 16 cartesian products of open sets in X_1 with open sets in X_2 and all the sets which can be obtained from them by taking unions and intersections. In total, these are 40 sets -- namely: {}, {(1,1)}, {(1,2)}, {(1,1),(1,2)}, {(1,1),(1,3)}, {(1,1),(2,1)}, {(1,2),(2,2)}, {(1,1),(1,2),(1,3)}, {(1,1),(1,2),(2,1)}, {(1,1),(1,2),(2,2)}, {(1,1),(1,3),(2,1)}, {(1,1),(2,1),(3,1)}, {(1,2),(2,2),(3,2)}, {(1,1),(1,2),(1,3),(2,1)}, {(1,1),(1,2),(1,3),(2,2)}, {(1,1),(1,2),(2,1),(2,2)}, {(1,1),(1,2),(2,1),(3,1)}, {(1,1),(1,2),(2,2),(3,2)}, {(1,1),(1,3),(2,1),(2,3)}, {(1,1),(1,3),(2,1),(3,1)}, {(1,1),(1,2),(1,3),(2,1),(2,2)}, {(1,1),(1,2),(1,3),(2,1),(2,3)}, {(1,1),(1,2),(1,3),(2,1),(3,1)}, {(1,1),(1,2),(1,3),(2,2),(3,2)}, {(1,1),(1,2),(2,1),(2,2),(3,1)}, {(1,1),(1,2),(2,1),(2,2),(3,2)}, {(1,1),(1,3),(2,1),(2,3),(3,1)}, {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)}, {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)}, {(1,1),(1,2),(1,3),(2,1),(2,2),(3,2)}, {(1,1),(1,2),(1,3),(2,1),(2,3),(3,1)}, {(1,1),(1,2),(2,1),(2,2),(3,1),(3,2)}, {(1,1),(1,3),(2,1),(2,3),(3,1),(3,3)}, {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1)}, {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,2)}, {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(3,2)}, {(1,1),(1,2),(1,3),(2,1),(2,3),(3,1),(3,3)}, {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2)}, {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,3)}, {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)} 25. The sets are finite and have the same cardinality, so an identification map must be an homeomorphism. However, the topological spaces X_1 and X_2 are not homeomorphic, since they have different numbers of open sets with one element. So there are no identification maps from X_1 to X_2. 26. The open sets are the images of the residue classes under q. These are just the residue classes which are subsets of r+mZ. The identification map q is a bijection, hence a homeomorphism. So the topologies are homeomorphic. 27. The trivial topology, i.e. the one in which the only open sets are the empty set and X itself. 28. All are compact, since they are finite. 29. 1. Not compact. -- The open cover consisting of the one-element subsets has no finite subcover. 2. Not compact. -- The open cover Z = 0(2) U 1(4) U 3(8) U 7(16) U ... has no finite subcover. 3. Compact. -- Given an open cover, construct a finite subcover as follows: take one set in the given cover, and cover the points in the complement one-by-one by adding at most one set per point. 4. Not compact. -- E.g. an open cover consisting of open discs of a fixed radius does not possess a finite subcover. 5. Compact, by Theorem 7.16. 6. Not compact. -- The open cover Q = S_1 U S_2 U S_3 U ... does not possess a finite subcover, since S_1 < S_2 < ... and Q is not equal to one of the S_n. 30. Every such function is bounded, by Theorem 7.16 and Lemma 7.5. 31. No. -- Counterexample: infinite discrete metric space. 32. No. -- Counterexample: R. 33. By the Triangle Inequality, the diameter is bounded below by d(1,6) = 6. It is bounded above by half of the circumference of the hexagon 1-2-3-4-5-6-1, which is (1+2+3+4+5+6)/2 = 10.5. It is straightforward to check that all diameters within these bounds can be realised. So we have 6 <= diam(X) <= 10. 34. 1. diam([0,1]^4) = sqrt(1^2+1^2+1^2+1^2) = sqrt(4) = 2. 2. We have d(x,y) = sum|x_i-y_i| <= sum|x_i| + sum|y_i| <= 1 + 1 = 2, hence the diameter of the set is 2. 35. 1. The problem is that the "Euclidean metric" on R^\infty, respectively S, is actually not a metric, since there are points with infinite distance. If one permits points with infinite distance ("extended metric"), the set S is not bounded. 2. The set S is not totally bounded, since any two of the infinitely many corners of the hypercube S have distance 1, which implies that a set with diameter less than 1 can at most cover one corner.